Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
Molar mass of water 18g/mol
Number of mols = 50.0g/18g/mol =2.78 mol
Heat absorbed = 40.7 kj/mol * 2.78 mol = 113.1 kj.
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use variable
1K₂MnF₆ + aSbF₅⇒ bKSbF₆ + cMnF₃ + dF₂
K, left=2,right=b⇒b=2
Mn, left=1, right=c⇒c=1
Sb, left=a, right=b⇒a=b=2
F, left=6.1+5a, right=6b+3c+2d
equation:
6+5(2) = 6(2)+3(1)+2d
16=15+2d
1=2d
d=0.5
So the reaction equation becomes:
1K₂MnF₆ + 2SbF₅⇒ 2KSbF₆ + 1MnF₃ + 0.5F₂ x2
2K₂MnF₆ + 4SbF₅⇒ 4KSbF₆ + 2MnF₃ + F₂
